On (04 Nov 96) James Di Paulo wrote to Ian Woofenden...
 JD> Another...  If one buys a 10KW wind generator, assume the
 JD> impossible, 100% efficiency, is this equivalent to 300KWH per
 JD> month? (10K X 30 days.) If not, what is the monthly output?
 JD> > 10kW is the _maximum_ output. That means if the wind speed
 JD> > is at the max rated value, it will put out 10 kW (watts are
 JD> > an instantaneous measurement of the RATE of power use). If
 JD> > it ran at the speed for an hour it would put out 10 kWh. If
 JD> > it ran at that speed for 30 _hours_ it would put out 300
 JD> > kWh.
 JD> That is what I need. Nowhere does it say that the 10KW is *per
 JD> hour*.
kW is a RATE. A watt is a RATE of power use. 
I agree that it does not _explicitly_ say that every time the term
is used. I think it's too bad that the terms "watt" and "amp" do not
_sound_ like rates. But they are just as much rates as "miles per
hour", as I understand it.
 JD> > The "monthly output" depends entirely upon your wind speeds.
 JD> > Get some average wind speed data, look at the machine's
 JD> > power curves and you could make a seat-of-the-pants guess at
 JD> > the monthly output.
 JD> Of course, but if one does not know how much the damn machine
 JD> is capable of producing in a *given amount of time*, the
 JD> graphs and curves are valueless. 
If you are talking about maximum output, that's easy. 
10 kW X 1 hour = 10 kWh.
It's getting a handle on what your available wind is that's tough...
Take it easy, but take it,
                        Ian
... I'm out of bed and dressed.  What more do you want?                       
 
... Us poor folks have to buy the best because we can't afford to buy twice.
--- PPoint 1.96
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